I have the below files located at some location in RHEL machine.
temp_file2.txt
temp_file3.txt
Looking for a python script to find above files recursively in all directories(I used a wild card, but it didn't work), and print a message if the file exists or not.
The below code snippet returns Nothing
import glob
for filename in glob.iglob('*/*.txt', recursive=True):
print(filename)
It returns the file name if it exists only in the current working directory
import glob
for filename in glob.iglob('.txt', recursive=True):
print(filename)
This approach seems to have worked for me, using python3.6
import glob
for f in glob.iglob('./**/*.yml', recursive=True):
print(f)
I was also able to use os.getcwd() + '/**/*.yml'. It appears there must be a directory definition at the start of the glob.
Related
I have folder where I have multiple files. Out of this files I want to rename some of them. For example: PB report December21 North.xlsb, PB report November21 North.xslb and so on. They all have a same start - PB report. I would like to change their name and leave only PB report and Month. For example PB report December.
I have tried this code:
import os
path = r'C://Users//greencolor//Desktop//Autoreport//Load_attachments//'
for filename in os.listdir(path):
if filename.startswith("PB report"):
os.rename(filename, filename[:-8])
-8 indicates that I want to split the name from the end on the 8th character
I get this error:
FileNotFoundError: [WinError 2] The system cannot find the file specified
Any suggestion?
You need the path when renaming file with os.rename:
Replace:
os.rename(filename, filename[:-8])
With:
filename_part, extension = os.path.splitext(filename)
os.rename(path+filename, path+filename_part[:-8]+extension)
The problem is likely that it cannot find the file because the directory is not specified. You need to add the path to the file name:
import os
path = r'C://Users//greencolor//Desktop//Autoreport//Load_attachments//'
for filename in os.listdir(path):
if filename.startswith("PB report"):
os.rename(os.path.join(path, filename), os.path.join(path, filename[:-8]))
This is a classic example of how working with os/os.path to manipulate paths is just not convenient. This is why pathlib exists. By treating paths as objects, rather than strings everything becomes more sensible. By using a combination of path.iterdir() and path.rename() you can achieve what you want like:
from pathlib import Path
path = Path(r'your path')
for file in path.iterdir():
if file.name.startswith("PB report"):
file.rename(file.with_stem(file.stem[:-8]))
Note that stem means the filename without the extension and that with_stem was added in Python 3.9. So for older versions you can still use with_name:
file.rename(file.with_name(file.stem[:-8] + file.suffix))
Where suffix is the extension of the file.
I have ~60 subdirectories in a single directory. Each of these contain thousands of files, but they all contain a file named test_all_results.txt.
What I would like to do is to rename each test_all_results.txt file so that it now has the name:
foldername_all_results.txt
What is the best way to do this?
Easily accomplished using Python os interface.
Assuming you are currently in the main directory:
import os
#get a list of all sub directories
subdir = os.listdir()
for dir in subdir:
if os.path.isdir(dir): #check if directory
os.chdir(dir) #move to sub directory
os.rename('test_all_results.txt', 'foldername_all_results.txt')
os.chdir('..') #return to main directory
Using python in Linux, make this:
import os
os.system("mv old_name.txt new_name.txt")
You can automatize with a loop, renaming all filenames.
You can do:
(change your code accordingly)
import os
# current directory is the target
direct = "."
for path, dirs, files in os.walk(direct):
for f in files:
if os.path.splitext(f)[0] == "test_all_results.txt":
os.rename(os.path.join(path, f), os.path.join(path, "foldername_all_results.txt"))
There's an answer that tells you to use the os.system() method, if you do decide to call Linux commands from Python, I'd advise that you use the subprocess module instead.
Here's how you'd run the mv command with two arguments using subprocess.call:
import subprocess
subprocess.call(["mv", "filename.txt", "new-name.txt"])
INFO: here's an old (but relevant) article that explains why it's dangerous to use these methods.
Good luck.
I am playing around with some python scripts and I ran into a problem with the script I'm writing. It's supposed to find all the files in a folder that meets the criteria and then delete it. However, it finds the files, but at the time of deleting the file, it says that the file is not found.
This is my code:
import os
for filename in os.listdir('C:\\New folder\\'):
if filename.endswith(".rdp"):
os.unlink(filename)
And this is the error I get after running it:
FileNotFoundError: [WinError 2] The system cannot find the file specified:
Can somebody assist with this?
os.unlink takes the path to the file, not only its filename. Try pre-pending your filename with the dirname. Like this
import os
dirname = 'C:\\New folder\\'
for filename in os.listdir(dirname):
if filename.endswith(".rdp"):
# Add your "dirname" to the file path
os.unlink(dirname + filename)
You could alternatively use os.walk, however it might go deeper than you want:
import os
for root, sub, file in os.walk("/media/"):
if file.endswith(".rdp"):
os.unlink(f'{root}/{file}')
I wanted to ask the forum how can I delete multiple files in a folder using Python. I tried using the import os module along with os.unlink() module, but it doesn't work. Any help would be greatly appreciated.
Most likely it's because the filename used in the os.unlink(filename) isn't a full path to the file (os.listdir() just returns a sequence of filenames). You probably need to use os.path.join() to prefix the 'c:\\users\\user1' folder to it first.
Something along these lines:
import os
folder = 'c:\\users\\user1\\Pictures'
for filename in os.listdir(folder):
if filename.endswith('.txt'):
os.unlink(os.path.join(folder, filename))
There is an mkv file in a folder named "export". What I want to do is to make a python script which fetches the file name from that export folder.
Let's say the folder is at "C:\Users\UserName\Desktop\New_folder\export".
How do I fetch the name?
I tried using this os.path.basename and os.path.splitext .. well.. didn't work out like I expected.
os.path implements some useful functions on pathnames. But it doesn't have access to the contents of the path. For that purpose, you can use os.listdir.
The following command will give you a list of the contents of the given path:
os.listdir("C:\Users\UserName\Desktop\New_folder\export")
Now, if you just want .mkv files you can use fnmatch(This module provides support for Unix shell-style wildcards) module to get your expected file names:
import fnmatch
import os
print([f for f in os.listdir("C:\Users\UserName\Desktop\New_folder\export") if fnmatch.fnmatch(f, '*.mkv')])
Also as #Padraic Cunningham mentioned as a more pythonic way for dealing with file names you can use glob module :
map(path.basename,glob.iglob(pth+"*.mkv"))
You can use glob:
from glob import glob
pth ="C:/Users/UserName/Desktop/New_folder/export/"
print(glob(pth+"*.mkv"))
path+"*.mkv" will match all the files ending with .mkv.
To just get the basenames you can use map or a list comp with iglob:
from glob import iglob
print(list(map(path.basename,iglob(pth+"*.mkv"))))
print([path.basename(f) for f in iglob(pth+"*.mkv")])
iglob returns an iterator so you don't build a list for no reason.
I assume you're basically asking how to list files in a given directory. What you want is:
import os
print os.listdir("""C:\Users\UserName\Desktop\New_folder\export""")
If there's multiple files and you want the one(s) that have a .mkv end you could do:
import os
files = os.listdir("""C:\Users\UserName\Desktop\New_folder\export""")
mkv_files = [_ for _ in files if _[-4:] == ".mkv"]
print mkv_files
If you are searching for recursive folder search, this method will help you to get filename using os.walk, also you can get those file's path and directory using this below code.
import os, fnmatch
for path, dirs, files in os.walk(os.path.abspath(r"C:/Users/UserName/Desktop/New_folder/export/")):
for filename in fnmatch.filter(files, "*.mkv"):
print(filename)
You can use glob
import glob
for file in glob.glob('C:\Users\UserName\Desktop\New_folder\export\*.mkv'):
print(str(file).split('\')[-1])
This will list out all the files having extention .mkv as
file.mkv, file2.mkv and so on.
From os.walk you can read file paths as a list
files = [ file_path for _, _, file_path in os.walk(DIRECTORY_PATH)]
for file_name in files[0]: #note that it has list of lists
print(file_name)